How Define Delta Function

I have a problem about calculating with Delta Function. I am trying write matlab code for these function.

I wrote the following code for this function.

n = -5:1:7;

x = delta(n+1) - delta(n) + unit(n+1) - unit(n-2);

stem(n,x,'fill');
axis([-6 8 -1.5 1.5])
xlabel('n')
ylabel('x[n]')
grid

But I am getting the following error.

I don't now how can ı write the delta function in this way ,

function y = unit(x);
y = zeros(size(x));
y(x>0) = 1;
end

Can you help me write the delta function as the 'unit' function you see above? Thank you for your helping.

Answer 1

[Star Strider]

If you have the Symbolic Math Toolbox, use the dirac function. It can be used with non-symbolic arguments as well.

[Walter Roberson]

dirac() is also available for double() and single() without symbolic toolbox.

[Kutlu Yigitturk]

When I write the code like this,

n = -5:1:7;

x = dirac(n+1) - dirac(n) + unit(n+1) - unit(n-2);

subplot(2,2,3);stem(n,x,'fill');
axis([-6 8 -1.5 1.5])
xlabel('n')
ylabel('x[n]')
grid

I get a graph like this,

But my main goal is to get this graph, what can I do about it?

[Walter Roberson]

That is not a true Dirac δ function.

delta = @(x) x==0;
unit = @(x) x>=0;
n = -5:1:7;
x = delta(n+1) - delta(n) + unit(n+1) - unit(n-2);
stem(n,x,'fill');
axis([-6 8 -1.5 2])
xlabel('n')
ylabel('x[n]')
grid

[Kutlu Yigitturk]

Thank you for your help Mr. Roberson. I got the result I wanted.

Answer 2

[Carlos M. Velez S.]

The examples above refer to the Kronecker delta function for discrete signals. If you want to apply the Dirac delta function in simulation to continuous-time systems, the following code is enough:

function y = delta_dirac(u)
[n,m] = size(u);
if max(n,m) ==1
    dt = 1e-6; % Define a small time increment for the delta function
else
    dt = u(2) - u(1);
end
y = zeros(n,m);
for i=1:max(m,n)
    if u(i) == 0
        y(i) = 1/dt;
    else
        y(i) = 0;
    end
end